Ways to sum to N using Natural Numbers up to K with repetitions allowed

Given two integers N and K, the task is to find the total number of ways of representing N as the sum of positive integers in the range [1, K], where each integer can be chosen multiple times.

Examples:

Input: N = 8, K = 2
Output: 5
Explanation: All possible ways of representing N as sum of positive integers less than or equal to K are:

1. {1, 1, 1, 1, 1, 1, 1, 1}, the sum is 8.
2. {2, 1, 1, 1, 1, 1, 1}, the sum is 8.
3. {2, 2, 1, 1, 1, 1}, the sum is 8.
4. 2, 2, 2, 1, 1}, the sum is 8.
5. {2, 2, 2, 2}}, the sum is 8.

Therefore, the total number of ways is 5.

Input: N = 2, K = 2
Output: 2

Naive Approach: The simplest approach to solve the given problem is to generate all possible combinations of choosing integers over the range [1, K] and count those combinations whose sum is N. 

Implementaion : 

//C++ program for above approach

#include <bits/stdc++.h>
using namespace std;


// Function that return Ways to sum to N using
// Natural Numbers up to K with repetitions allowed
int NumberOfways(int N, int K) {
    
    // Base case
    if (N == 0) return 1;
    if (N < 0 || K <= 0) return 0;
    
    // including and not including K in sum
    return NumberOfways(N - K, K) + NumberOfways(N, K - 1);
}


// Driver code
int main() {
    int N = 8;
    int K = 2;
    
    // function call
    cout << NumberOfways(N, K) << endl;
    return 0;
}

// this code is contributed by bhardwajji

import java.util.*;

class Main {
    // Function that return Ways to sum to N using
    // Natural Numbers up to K with repetitions allowed
    public static int NumberOfways(int N, int K)
    {
        // Base case
        if (N == 0)
            return 1;
        if (N < 0 || K <= 0)
            return 0;

        // including and not including K in sum
        return NumberOfways(N - K, K)
            + NumberOfways(N, K - 1);
    }

    // Driver code
    public static void main(String[] args)
    {
        int N = 8;
        int K = 2;

        // function call
        System.out.println(NumberOfways(N, K));
    }
}

# Function that returns the number of ways to sum
# to N using natural numbers up to K with repetitions allowed


def number_of_ways(N, K):

    # Base case
    if N == 0:
        return 1
    if N < 0 or K <= 0:
        return 0

    # Including and not including K in sum
    return number_of_ways(N - K, K) + number_of_ways(N, K - 1)


# Driver code
if __name__ == '__main__':
    N = 8
    K = 2

    # Function call
    print(number_of_ways(N, K))

// Function that return Ways to sum to N using
// Natural Numbers up to K with repetitions allowed
function numberOfWays(N, K) {
    
    // Base case
    if (N == 0) return 1;
    if (N < 0 || K <= 0) return 0;
    
    // including and not including K in sum
    return numberOfWays(N - K, K) + numberOfWays(N, K - 1);
}


// Driver code
let N = 8;
let K = 2;
    
// function call
console.log(numberOfWays(N, K));

using System;

class MainClass {
    // Function that return Ways to sum to N using
    // Natural Numbers up to K with repetitions allowed
    static int NumberOfWays(int N, int K)
    {
        // Base case
        if (N == 0)
            return 1;
        if (N < 0 || K <= 0)
            return 0;

        // including and not including K in sum
        return NumberOfWays(N - K, K)
            + NumberOfWays(N, K - 1);
    }

    // Driver code
    static void Main()
    {
        int N = 8;
        int K = 2;

        // function call
        Console.WriteLine(NumberOfWays(N, K));
    }
}
// This code is contributed by user_dtewbxkn77n

5

Time Complexity: O(K^N)
Auxiliary Space: O(1)

Efficient Approach: The above approach has Overlapping Subproblems and an Optimal Substructure. Hence, in order to optimize, Dynamic Programming is needed to be performed based on the following observations:

• Considering dp[i] stores the total number of ways for representing i as the sum of integers lying in the range [1, K], then the transition of states can be defined as:
  • For i in the range [1, K] and for every j in the range [1, N]
  • The value of dp[j] is equal to (dp[j]+ dp[j - i]), for all j ? i.

Follow the steps below to solve the problem:

• Initialize an array, say dp[], with all elements as 0, to store all the recursive states.
• Initialize dp[0] as 1.
• Now, iterate over the range [1, K] using a variable i and perform the following steps: 
  
  • Iterate over the range [1, N], using a variable j, and update the value of dp[j] as dp[j]+ dp[j - i], if j ? i.
• After completing the above steps, print the value of dp[N] as the result.

Below is the implementation of the above approach:

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the total number of
// ways to represent N as the sum of
// integers over the range [1, K]
int NumberOfways(int N, int K)
{

    // Initialize a list
    vector<int> dp(N + 1, 0);
  
    // Update dp[0] to 1
    dp[0] = 1;

    // Iterate over the range [1, K + 1]
    for (int row = 1; row < K + 1; row++)
    {

        // Iterate over the range [1, N + 1]
        for (int col = 1; col < N + 1; col++)
        {

            // If col is greater
            // than or equal to row
            if (col >= row)
              
                // Update current
                // dp[col] state
                dp[col] = dp[col] + dp[col - row];
          }
    }

    // Return the total number of ways
    return(dp[N]);
}

// Driver Code
int main()
{
  int N = 8;
  int K = 2;

  cout << (NumberOfways(N, K));
}

// This code is contributed by mohit kumar 29.

// Java program for the above approach
import java.util.*;

class GFG{
    
// Function to find the total number of
// ways to represent N as the sum of
// integers over the range [1, K]
static int NumberOfways(int N, int K)
{
    
    // Initialize a list
    int[] dp = new int[N + 1];
  
    // Update dp[0] to 1
    dp[0] = 1;

    // Iterate over the range [1, K + 1]
    for(int row = 1; row < K + 1; row++)
    {

        // Iterate over the range [1, N + 1]
        for(int col = 1; col < N + 1; col++)
        {
            
            // If col is greater
            // than or equal to row
            if (col >= row)
              
                // Update current
                // dp[col] state
                dp[col] = dp[col] + dp[col - row];
          }
    }

    // Return the total number of ways
    return(dp[N]);
}

// Driver code
public static void main(String[] args)
{
    
    // Given inputs
    int N = 8;
    int K = 2;
    
    System.out.println(NumberOfways(N, K));
}
}

// This code is contributed by offbeat

# Python program for the above approach

# Function to find the total number of
# ways to represent N as the sum of
# integers over the range [1, K]
def NumberOfways(N, K):
  
    # Initialize a list
    dp = [0] * (N + 1)
    
    # Update dp[0] to 1
    dp[0] = 1
    
    # Iterate over the range [1, K + 1]
    for row in range(1, K + 1):
      
        # Iterate over the range [1, N + 1]
        for col in range(1, N + 1):
          
            # If col is greater
            # than or equal to row
            if (col >= row):
              
                # Update current
                # dp[col] state
                dp[col] = dp[col] + dp[col - row]
                
                
    # Return the total number of ways
    return(dp[N])

# Driver Code

N = 8
K = 2

print(NumberOfways(N, K))

<script>
// Javascript implementation for the above approach

// Function to find the total number of
// ways to represent N as the sum of
// integers over the range [1, K]
function NumberOfways(N, K)
{
    
    // Initialize a list
    let dp = Array.from({length: N +1}, (_, i) => 0);
  
    // Update dp[0] to 1
    dp[0] = 1;

    // Iterate over the range [1, K + 1]
    for(let row = 1; row < K + 1; row++)
    {

        // Iterate over the range [1, N + 1]
        for(let col = 1; col < N + 1; col++)
        {
            
            // If col is greater
            // than or equal to row
            if (col >= row)
              
                // Update current
                // dp[col] state
                dp[col] = dp[col] + dp[col - row];
          }
    }

    // Return the total number of ways
    return(dp[N]);
}

    // Driver Code
    
    // Given inputs
    let N = 8;
    let K = 2;
    
    document.write(NumberOfways(N, K));

</script>

// C# program for the above approach
using System;
class GFG 
{
  
    // Function to find the total number of
    // ways to represent N as the sum of
    // integers over the range [1, K]
    static int NumberOfways(int N, int K)
    {

        // Initialize a list
        int[] dp = new int[(N + 1)];

        // Update dp[0] to 1
        dp[0] = 1;

        // Iterate over the range [1, K + 1]
        for (int row = 1; row < K + 1; row++) {

            // Iterate over the range [1, N + 1]
            for (int col = 1; col < N + 1; col++) {

                // If col is greater
                // than or equal to row
                if (col >= row)

                    // Update current
                    // dp[col] state
                    dp[col] = dp[col] + dp[col - row];
            }
        }

        // Return the total number of ways
        return (dp[N]);
    }

    // Driver Code
    public static void Main()
    {
        int N = 8;
        int K = 2;

        Console.WriteLine(NumberOfways(N, K));
    }
}

// This code is contributed by ukasp.

5

Time Complexity: O(N * K)
Auxiliary Space: O(N)